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Lesson 12 of 16
  • Year 10
  • Higher

Solving quadratic equations by factorising where rearrangement is required

I can solve quadratic equations algebraically by factorising where rearrangement is required.

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Lesson 12 of 16
New
New
  • Year 10
  • Higher

Solving quadratic equations by factorising where rearrangement is required

I can solve quadratic equations algebraically by factorising where rearrangement is required.

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Share activities with pupils

These resources will be removed by end of Summer Term 2025.

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These resources were created for remote use during the pandemic and are not designed for classroom teaching.

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Lesson details

Key learning points

  1. In order for factorising to be a valid method, the quadratic must equal zero.
  2. To achieve a product of zero, one of the binomial expressions evaluate to zero.
  3. Either expression could be zero so you must find the solution for each expression.
  4. If the quadratic were not equal to zero, you would have uncertainty about what each expression must equal.

Keywords

  • Factorise - To factorise is to express a term as the product of its factors.

  • Solution (equality) - A solution to an equality with one variable is a value for the variable which, when substituted, maintains the equality between the expressions.

Common misconception

Pupils may try to factorise and solve without rearranging first.

Being able to put the factors equal to zero only makes sense if we are using the property that if two values multiply to zero, one of them is zero.


To help you plan your year 10 maths lesson on: Solving quadratic equations by factorising where rearrangement is required, download all teaching resources for free and adapt to suit your pupils' needs...

In context, it is very rare that equations will be equal to zero to start. The ability to rearrange is essential to solve most problems. Pupils should be able to access most problems with quadratics at the end of this lesson.
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Licence

This content is © Oak National Academy Limited (2025), licensed on Open Government Licence version 3.0 except where otherwise stated. See Oak's terms & conditions (Collection 2).

Lesson video

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Prior knowledge starter quiz

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6 Questions

Q1.
Factorise the expression $$x^2 + 2x - 15$$

Correct answer: $$(x - 3)(x + 5)$$
$$(x + 3)(x - 5)$$
$$(x + 15)(x - 1)$$

Q2.
Which of these is a solution for: $$(x - 4)(x + 2) = 0$$

$$x = 0$$
Correct answer: $$x = 4$$
$$x = -4$$
$$x = 2$$

Q3.
Which of these is a solution for: $$(x + 2)(x - 5) = 0$$

$$x = 2$$
$$x = -5$$
Correct answer: $$x = 5$$
$$x = 0$$

Q4.
Which of these is a solution for: $$(x + 9)(x + 3) = 0$$

$$x = 9$$
$$x = 3$$
Correct answer: $$x = -3$$
Correct answer: $$x = -9$$

Q5.
Which of these is a solution for: $$(x - 6)^2 = 0$$

$$x = -6$$
Correct answer: $$x = 6$$
$$x = 0$$

Q6.
Factorise the expression $$7x^2 + 42x - 49$$

$$7(x + 1)(x - 7)$$
$$7(x - 1)(x - 7)$$
Correct answer: $$7(x - 1)(x + 7)$$

Assessment exit quiz

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6 Questions

Q1.
Find the value of $$x$$ where: $$x^2 = 6x - 9$$

Correct Answer: 3

Q2.
Find the value of $$x$$ where: $$x^2 = 4x - 4$$

Correct Answer: 2

Q3.
Find the values of $$x$$ where: $$5x = x^2 - 6$$

$$x = 1, x = -6$$
$$x = -1, x = -6$$
Correct answer: $$x = -1, x = 6$$

Q4.
Solve the equation: $$x^2 = 8x + 9$$

$$x = -1, x = -9$$
$$x = 1, x = -9$$
Correct answer: $$x = -1, x = 9$$

Q5.
Solve the equation: $$2x^2 - 3x = 9$$

$$x = -3, x = -1.5$$
Correct answer: $$x = 3, x = -1.5$$
$$x = -3, x = 1.5$$

Q6.
Solve the equation: $$x^2 - 2 = x$$

Correct answer: $$x = 2, x = -1$$
$$x = -2, x = -1$$
$$x = 2, x = 1$$